Description of Ring and Ideal

First I'm sorry this is my first post and I don't how to entirely format math equations yet, but I'm trying to figure it out.

But I am having a difficult time understanding this specific case of a quotient ring. I know that generally speaking the product of elements of a quotient ring are

${(a + I)(b + I) = (ab + I)}$

where I is an ideal in the ring.

So for my specific question in the quotient ring R[x]/((x^2)-4) the problem states that each coset has the unique form ${a + bx + (x^2-4)}$ given that a and b are real numbers. The problem then asks to express the product of two of these cosets

${[a + bx + (x^2-4)] * [c + dx + (x^2-4)]}$

in the same form, that is, to find the product in the form ${e + fx + (x^2-4)}$ and write out what e and f are in terms of a,b,c, and d.

What I have so far is this:

I'm essentially taking ${(a + I)(b + I) = (ab + I)}$ to be

${(a + bx + (x^2-4))(c + dx + (x^2-4))}$

and then

${(a + bx)(c + dx) = ac + x(ad + cb) + bdx^2}$

But from my understanding of how the quotient ring works is that since the principal ideal is ${(x^2-4)}$ then that leads to an additional condition that we have ${x^2-4 = 0}$ which implies ${x^2 = 4}$. So then my equation above becomes

${ac + x(ad + cb) + bdx^2 = ac + x(ad + cb) + 4bd}$

and so in the form the question is asking me it would then be that for the product of the two cosets in terms of a,b,c, and d we have that

$${e + fx + (x^2-4)}$$ $${e = 4bd + ac}$$ $${f = ad + cb}$$

Is this the correct way of going about multiplying two elements of this quotient ring? Or am I entirely missing the process here?

  • $\begingroup$ What do you doubt about your work? $\endgroup$ – Bill Dubuque Dec 7 '20 at 9:24
  • $\begingroup$ Looks good to me. $\endgroup$ – Gerry Myerson Dec 7 '20 at 9:37
  • $\begingroup$ I was both unsure of only taking the first two expressions and leaving out the (x^2-4) and having the condition of x^2 = 4. I have been having an incredibly difficult time understand abstract algebra and wrapping my head around the concepts. $\endgroup$ – Z Ham Dec 7 '20 at 19:08

You can't do $x^2=4$, only “sort of”. What you do know is that $$ x^2-4\in(x^2-4)=I $$

and therefore $x^2+I=4+I$.

So when you multiply $$ (a+bx+I)(c+dx+I)=ac+(ad+bc)x+bdx^2+I $$ you can use $bdx^2+I=4bd+I$ and you end up with $$ (a+bx+I)(c+dx+I)=(ac+4bd)+(ad+bc)x+I $$

  • $\begingroup$ Thank you for elaborating on that concept, I wasn't sure why it was the way it was. $\endgroup$ – Z Ham Dec 7 '20 at 19:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.